\(\int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) [404]

Optimal result

Integrand size = 29, antiderivative size = 141 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}+\frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d} \] Output:

-a^3*(a^2+b^2)*ln(a+b*sinh(d*x+c))/b^6/d+a^2*(a^2+b^2)*sinh(d*x+c)/b^5/d-1 
/2*a*(a^2+b^2)*sinh(d*x+c)^2/b^4/d+1/3*(a^2+b^2)*sinh(d*x+c)^3/b^3/d-1/4*a 
*sinh(d*x+c)^4/b^2/d+1/5*sinh(d*x+c)^5/b/d
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {60 a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6}-\frac {60 a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5}+\frac {30 a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{b^4}-\frac {20 \left (a^2+b^2\right ) \sinh ^3(c+d x)}{b^3}+\frac {15 a \sinh ^4(c+d x)}{b^2}-\frac {12 \sinh ^5(c+d x)}{b}}{60 d} \] Input:

Integrate[(Cosh[c + d*x]^3*Sinh[c + d*x]^3)/(a + b*Sinh[c + d*x]),x]
 

Output:

-1/60*((60*a^3*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/b^6 - (60*a^2*(a^2 + 
b^2)*Sinh[c + d*x])/b^5 + (30*a*(a^2 + b^2)*Sinh[c + d*x]^2)/b^4 - (20*(a^ 
2 + b^2)*Sinh[c + d*x]^3)/b^3 + (15*a*Sinh[c + d*x]^4)/b^2 - (12*Sinh[c + 
d*x]^5)/b)/d
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 26, 3316, 26, 27, 522, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sinh ^3(c+d x) \cosh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \sin (i c+i d x)^3 \cos (i c+i d x)^3}{a-i b \sin (i c+i d x)}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\cos (i c+i d x)^3 \sin (i c+i d x)^3}{a-i b \sin (i c+i d x)}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle -\frac {i \int \frac {i \sinh ^3(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^3 d}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\int \frac {\sinh ^3(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b^3 \sinh ^3(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^6 d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {\int \left (b^4 \sinh ^4(c+d x)-a b^3 \sinh ^3(c+d x)+b^2 \left (a^2+b^2\right ) \sinh ^2(c+d x)-a b \left (a^2+b^2\right ) \sinh (c+d x)+a^2 \left (a^2+b^2\right )-\frac {a^3 \left (a^2+b^2\right )}{a+b \sinh (c+d x)}\right )d(b \sinh (c+d x))}{b^6 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{2} a b^2 \left (a^2+b^2\right ) \sinh ^2(c+d x)+a^2 b \left (a^2+b^2\right ) \sinh (c+d x)+\frac {1}{3} b^3 \left (a^2+b^2\right ) \sinh ^3(c+d x)-a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-\frac {1}{4} a b^4 \sinh ^4(c+d x)+\frac {1}{5} b^5 \sinh ^5(c+d x)}{b^6 d}\)

Input:

Int[(Cosh[c + d*x]^3*Sinh[c + d*x]^3)/(a + b*Sinh[c + d*x]),x]
 

Output:

(-(a^3*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]]) + a^2*b*(a^2 + b^2)*Sinh[c + 
d*x] - (a*b^2*(a^2 + b^2)*Sinh[c + d*x]^2)/2 + (b^3*(a^2 + b^2)*Sinh[c + d 
*x]^3)/3 - (a*b^4*Sinh[c + d*x]^4)/4 + (b^5*Sinh[c + d*x]^5)/5)/(b^6*d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 152.48 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {\frac {\sinh \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sinh \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sinh \left (d x +c \right )^{3}}{3}+\frac {b^{4} \sinh \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sinh \left (d x +c \right )^{2}}{2}-\frac {a \,b^{3} \sinh \left (d x +c \right )^{2}}{2}+a^{4} \sinh \left (d x +c \right )+a^{2} b^{2} \sinh \left (d x +c \right )}{b^{5}}-\frac {a^{3} \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{6}}}{d}\)	\(144\)
default	\(\frac {\frac {\frac {\sinh \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sinh \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sinh \left (d x +c \right )^{3}}{3}+\frac {b^{4} \sinh \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sinh \left (d x +c \right )^{2}}{2}-\frac {a \,b^{3} \sinh \left (d x +c \right )^{2}}{2}+a^{4} \sinh \left (d x +c \right )+a^{2} b^{2} \sinh \left (d x +c \right )}{b^{5}}-\frac {a^{3} \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{6}}}{d}\)	\(144\)
risch	\(-\frac {{\mathrm e}^{d x +c}}{16 b d}+\frac {{\mathrm e}^{-d x -c}}{16 b d}+\frac {{\mathrm e}^{5 d x +5 c}}{160 b d}+\frac {a^{3} x}{b^{4}}-\frac {{\mathrm e}^{-3 d x -3 c}}{96 b d}-\frac {a \,{\mathrm e}^{-4 d x -4 c}}{64 b^{2} d}-\frac {a \,{\mathrm e}^{4 d x +4 c}}{64 b^{2} d}+\frac {3 \,{\mathrm e}^{d x +c} a^{2}}{8 b^{3} d}-\frac {3 \,{\mathrm e}^{-d x -c} a^{2}}{8 b^{3} d}+\frac {2 a^{3} c}{b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{4} d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{16 b^{2} d}-\frac {a \,{\mathrm e}^{2 d x +2 c}}{16 b^{2} d}+\frac {{\mathrm e}^{3 d x +3 c}}{96 b d}-\frac {{\mathrm e}^{-5 d x -5 c}}{160 b d}-\frac {{\mathrm e}^{-d x -c} a^{4}}{2 b^{5} d}-\frac {a^{3} {\mathrm e}^{-2 d x -2 c}}{8 b^{4} d}-\frac {{\mathrm e}^{-3 d x -3 c} a^{2}}{24 b^{3} d}+\frac {2 a^{5} c}{b^{6} d}-\frac {a^{5} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{6} d}+\frac {{\mathrm e}^{3 d x +3 c} a^{2}}{24 b^{3} d}-\frac {a^{3} {\mathrm e}^{2 d x +2 c}}{8 b^{4} d}+\frac {{\mathrm e}^{d x +c} a^{4}}{2 b^{5} d}+\frac {a^{5} x}{b^{6}}\)	\(437\)

Input:

int(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/b^5*(1/5*sinh(d*x+c)^5*b^4-1/4*a*sinh(d*x+c)^4*b^3+1/3*a^2*b^2*sinh 
(d*x+c)^3+1/3*b^4*sinh(d*x+c)^3-1/2*a^3*b*sinh(d*x+c)^2-1/2*a*b^3*sinh(d*x 
+c)^2+a^4*sinh(d*x+c)+a^2*b^2*sinh(d*x+c))-a^3*(a^2+b^2)/b^6*ln(a+b*sinh(d 
*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1660 vs. \(2 (133) = 266\).

Time = 0.11 (sec) , antiderivative size = 1660, normalized size of antiderivative = 11.77 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="fric 
as")
 

Output:

1/960*(6*b^5*cosh(d*x + c)^10 + 6*b^5*sinh(d*x + c)^10 - 15*a*b^4*cosh(d*x 
 + c)^9 + 15*(4*b^5*cosh(d*x + c) - a*b^4)*sinh(d*x + c)^9 + 10*(4*a^2*b^3 
 + b^5)*cosh(d*x + c)^8 + 5*(54*b^5*cosh(d*x + c)^2 - 27*a*b^4*cosh(d*x + 
c) + 8*a^2*b^3 + 2*b^5)*sinh(d*x + c)^8 + 960*(a^5 + a^3*b^2)*d*x*cosh(d*x 
 + c)^5 - 60*(2*a^3*b^2 + a*b^4)*cosh(d*x + c)^7 + 20*(36*b^5*cosh(d*x + c 
)^3 - 27*a*b^4*cosh(d*x + c)^2 - 6*a^3*b^2 - 3*a*b^4 + 4*(4*a^2*b^3 + b^5) 
*cosh(d*x + c))*sinh(d*x + c)^7 + 60*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x 
+ c)^6 + 20*(63*b^5*cosh(d*x + c)^4 - 63*a*b^4*cosh(d*x + c)^3 + 24*a^4*b 
+ 18*a^2*b^3 - 3*b^5 + 14*(4*a^2*b^3 + b^5)*cosh(d*x + c)^2 - 21*(2*a^3*b^ 
2 + a*b^4)*cosh(d*x + c))*sinh(d*x + c)^6 - 15*a*b^4*cosh(d*x + c) + 2*(75 
6*b^5*cosh(d*x + c)^5 - 945*a*b^4*cosh(d*x + c)^4 + 280*(4*a^2*b^3 + b^5)* 
cosh(d*x + c)^3 + 480*(a^5 + a^3*b^2)*d*x - 630*(2*a^3*b^2 + a*b^4)*cosh(d 
*x + c)^2 + 180*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c))*sinh(d*x + c)^5 
 - 6*b^5 - 60*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c)^4 + 10*(126*b^5*co 
sh(d*x + c)^6 - 189*a*b^4*cosh(d*x + c)^5 - 48*a^4*b - 36*a^2*b^3 + 6*b^5 
+ 70*(4*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 480*(a^5 + a^3*b^2)*d*x*cosh(d*x 
+ c) - 210*(2*a^3*b^2 + a*b^4)*cosh(d*x + c)^3 + 90*(8*a^4*b + 6*a^2*b^3 - 
 b^5)*cosh(d*x + c)^2)*sinh(d*x + c)^4 - 60*(2*a^3*b^2 + a*b^4)*cosh(d*x + 
 c)^3 + 20*(36*b^5*cosh(d*x + c)^7 - 63*a*b^4*cosh(d*x + c)^6 + 28*(4*a^2* 
b^3 + b^5)*cosh(d*x + c)^5 - 6*a^3*b^2 - 3*a*b^4 + 480*(a^5 + a^3*b^2)*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cosh(d*x+c)**3*sinh(d*x+c)**3/(a+b*sinh(d*x+c)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (133) = 266\).

Time = 0.05 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.13 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (15 \, a b^{3} e^{\left (-d x - c\right )} - 6 \, b^{4} - 10 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, {\left (2 \, a^{3} b + a b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} - 60 \, {\left (8 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} e^{\left (5 \, d x + 5 \, c\right )}}{960 \, b^{5} d} - \frac {{\left (a^{5} + a^{3} b^{2}\right )} {\left (d x + c\right )}}{b^{6} d} - \frac {15 \, a b^{3} e^{\left (-4 \, d x - 4 \, c\right )} + 6 \, b^{4} e^{\left (-5 \, d x - 5 \, c\right )} + 60 \, {\left (8 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-d x - c\right )} + 60 \, {\left (2 \, a^{3} b + a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{960 \, b^{5} d} - \frac {{\left (a^{5} + a^{3} b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{6} d} \] Input:

integrate(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="maxi 
ma")
                                                                                    
                                                                                    
 

Output:

-1/960*(15*a*b^3*e^(-d*x - c) - 6*b^4 - 10*(4*a^2*b^2 + b^4)*e^(-2*d*x - 2 
*c) + 60*(2*a^3*b + a*b^3)*e^(-3*d*x - 3*c) - 60*(8*a^4 + 6*a^2*b^2 - b^4) 
*e^(-4*d*x - 4*c))*e^(5*d*x + 5*c)/(b^5*d) - (a^5 + a^3*b^2)*(d*x + c)/(b^ 
6*d) - 1/960*(15*a*b^3*e^(-4*d*x - 4*c) + 6*b^4*e^(-5*d*x - 5*c) + 60*(8*a 
^4 + 6*a^2*b^2 - b^4)*e^(-d*x - c) + 60*(2*a^3*b + a*b^3)*e^(-2*d*x - 2*c) 
 + 10*(4*a^2*b^2 + b^4)*e^(-3*d*x - 3*c))/(b^5*d) - (a^5 + a^3*b^2)*log(-2 
*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/(b^6*d)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.83 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {6 \, b^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} - 15 \, a b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4} + 40 \, a^{2} b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 40 \, b^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 120 \, a^{3} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 120 \, a b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 480 \, a^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 480 \, a^{2} b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{5}} - \frac {960 \, {\left (a^{5} + a^{3} b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{6}}}{960 \, d} \] Input:

integrate(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="giac 
")
 

Output:

1/960*((6*b^4*(e^(d*x + c) - e^(-d*x - c))^5 - 15*a*b^3*(e^(d*x + c) - e^( 
-d*x - c))^4 + 40*a^2*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + 40*b^4*(e^(d*x 
+ c) - e^(-d*x - c))^3 - 120*a^3*b*(e^(d*x + c) - e^(-d*x - c))^2 - 120*a* 
b^3*(e^(d*x + c) - e^(-d*x - c))^2 + 480*a^4*(e^(d*x + c) - e^(-d*x - c)) 
+ 480*a^2*b^2*(e^(d*x + c) - e^(-d*x - c)))/b^5 - 960*(a^5 + a^3*b^2)*log( 
abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/b^6)/d
 

Mupad [B] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.18 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\mathrm {e}}^{5\,c+5\,d\,x}}{160\,b\,d}-\frac {{\mathrm {e}}^{-5\,c-5\,d\,x}}{160\,b\,d}-\frac {a\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,b^2\,d}-\frac {a\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,b^2\,d}-\frac {\ln \left (2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^5+a^3\,b^2\right )}{b^6\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (8\,a^4+6\,a^2\,b^2-b^4\right )}{16\,b^5\,d}+\frac {a^3\,x\,\left (a^2+b^2\right )}{b^6}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (2\,a^3+a\,b^2\right )}{16\,b^4\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3+a\,b^2\right )}{16\,b^4\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (8\,a^4+6\,a^2\,b^2-b^4\right )}{16\,b^5\,d}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,\left (4\,a^2+b^2\right )}{96\,b^3\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (4\,a^2+b^2\right )}{96\,b^3\,d} \] Input:

int((cosh(c + d*x)^3*sinh(c + d*x)^3)/(a + b*sinh(c + d*x)),x)
 

Output:

exp(5*c + 5*d*x)/(160*b*d) - exp(- 5*c - 5*d*x)/(160*b*d) - (a*exp(- 4*c - 
 4*d*x))/(64*b^2*d) - (a*exp(4*c + 4*d*x))/(64*b^2*d) - (log(2*a*exp(d*x)* 
exp(c) - b + b*exp(2*c)*exp(2*d*x))*(a^5 + a^3*b^2))/(b^6*d) - (exp(- c - 
d*x)*(8*a^4 - b^4 + 6*a^2*b^2))/(16*b^5*d) + (a^3*x*(a^2 + b^2))/b^6 - (ex 
p(- 2*c - 2*d*x)*(a*b^2 + 2*a^3))/(16*b^4*d) - (exp(2*c + 2*d*x)*(a*b^2 + 
2*a^3))/(16*b^4*d) + (exp(c + d*x)*(8*a^4 - b^4 + 6*a^2*b^2))/(16*b^5*d) - 
 (exp(- 3*c - 3*d*x)*(4*a^2 + b^2))/(96*b^3*d) + (exp(3*c + 3*d*x)*(4*a^2 
+ b^2))/(96*b^3*d)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.02 \[ \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {6 e^{10 d x +10 c} b^{5}-15 e^{9 d x +9 c} a \,b^{4}+40 e^{8 d x +8 c} a^{2} b^{3}+10 e^{8 d x +8 c} b^{5}-120 e^{7 d x +7 c} a^{3} b^{2}-60 e^{7 d x +7 c} a \,b^{4}+480 e^{6 d x +6 c} a^{4} b +360 e^{6 d x +6 c} a^{2} b^{3}-60 e^{6 d x +6 c} b^{5}-960 e^{5 d x +5 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a^{5}-960 e^{5 d x +5 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a^{3} b^{2}+960 e^{5 d x +5 c} a^{5} d x +960 e^{5 d x +5 c} a^{3} b^{2} d x -480 e^{4 d x +4 c} a^{4} b -360 e^{4 d x +4 c} a^{2} b^{3}+60 e^{4 d x +4 c} b^{5}-120 e^{3 d x +3 c} a^{3} b^{2}-60 e^{3 d x +3 c} a \,b^{4}-40 e^{2 d x +2 c} a^{2} b^{3}-10 e^{2 d x +2 c} b^{5}-15 e^{d x +c} a \,b^{4}-6 b^{5}}{960 e^{5 d x +5 c} b^{6} d} \] Input:

int(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x)
 

Output:

(6*e**(10*c + 10*d*x)*b**5 - 15*e**(9*c + 9*d*x)*a*b**4 + 40*e**(8*c + 8*d 
*x)*a**2*b**3 + 10*e**(8*c + 8*d*x)*b**5 - 120*e**(7*c + 7*d*x)*a**3*b**2 
- 60*e**(7*c + 7*d*x)*a*b**4 + 480*e**(6*c + 6*d*x)*a**4*b + 360*e**(6*c + 
 6*d*x)*a**2*b**3 - 60*e**(6*c + 6*d*x)*b**5 - 960*e**(5*c + 5*d*x)*log(e* 
*(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b)*a**5 - 960*e**(5*c + 5*d*x)*log(e 
**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b)*a**3*b**2 + 960*e**(5*c + 5*d*x) 
*a**5*d*x + 960*e**(5*c + 5*d*x)*a**3*b**2*d*x - 480*e**(4*c + 4*d*x)*a**4 
*b - 360*e**(4*c + 4*d*x)*a**2*b**3 + 60*e**(4*c + 4*d*x)*b**5 - 120*e**(3 
*c + 3*d*x)*a**3*b**2 - 60*e**(3*c + 3*d*x)*a*b**4 - 40*e**(2*c + 2*d*x)*a 
**2*b**3 - 10*e**(2*c + 2*d*x)*b**5 - 15*e**(c + d*x)*a*b**4 - 6*b**5)/(96 
0*e**(5*c + 5*d*x)*b**6*d)